Maths-
General
Easy

Question

A 12% brine solution was mixed with a 16% brine solution to produce a 15% brine solution. How much of the 12% brine solution and how much of the 16% brine solution were used to produce 40 L of the 15% solution?

Hint:

let the 12 % brine solution be x L and the 16% brine solution mixed be y L
Total amount of mixed solution = amount of 12 % solution +amount of 16 % solution
= 40 L
Amount of salt in final brine solution = amount of salt in 15% brine solution
= amount of salt in 12% brine +  amount of salt in 16% brine

The correct answer is: 16%


    Ans :- 30 L of 12% brine solution and 10 L of 16% brine solution
    Explanation :-
    Step 1:-Frame system of linear equations by given condition
    Total amount of mixed solution = amount of 12 % solution +amount of 16 % solution
    x L plus y L equals 40 L not stretchy rightwards double arrow x plus y equals 40 — Eq1
    The amount of salt in 12% of x Litres brine solution = open parentheses 12 over 100 close parentheses straight x equals fraction numerator 3 x over denominator 25 end fraction
    The amount of salt in 16% of y Litres brine solution = open parentheses 10 over 100 close parentheses straight y equals fraction numerator 4 y over denominator 25 end fraction
    The amount of salt in 15% of  40 L of brine solution = open parentheses 15 over 100 close parentheses space cross times space 40 equals fraction numerator 15 space cross times space 10 over denominator 25 end fraction
    Amount of salt in final brine solution = amount of salt in 15% brine solution
    = amount of salt in 12% brine +  amount of salt in 16% brine
    We get, fraction numerator 3 x over denominator 25 end fraction plus fraction numerator 4 x over denominator 25 end fraction equals fraction numerator 15 ∗ 10 over denominator 25 end fraction not stretchy rightwards double arrow 3 x plus 4 y equals 150 — Eq2
    Step 2:- eliminate x to find the value of y .
    Doing (Eq2) -3(Eq1)  to eliminate x .
    left parenthesis 3 x plus 4 y right parenthesis minus 3 left parenthesis x plus y right parenthesis equals left parenthesis 150 right parenthesis minus 3 left parenthesis 40 right parenthesis
    3 x minus 3 x plus 4 y minus 3 y equals 150 minus 120
    y equals 30
    ∴y = 30
    Step 3:- find x by substituting the value of y in eq1.
    x plus y equals 40 not stretchy rightwards double arrow x plus 30 equals 40
    not stretchy rightwards double arrow x equals 40 minus 30 equals 10
    ∴x =10
    ∴30 L of 12% brine solution and 10 L of 16% brine solution is mixed to get 40 L of 15 % brine solution .

    Related Questions to study

    General
    Maths-

    Candies  cost $2 an ounce and have 500 calories in an ounce. Cookies cost $3 an ounce and have 400calories in an ounce. A mixture of candies and cookies costs $29 and has 4,800 calories. Create a system of equations to model this scenario. How many ounces of candies are there in the mixture?

    Ans :-The no. of ounces of candy is 4 ounces in the mixture .
    Explanation :-
    Given ,candies  cost $2 an ounce and have 500 calories in an ounce. Cookies cost $3 an ounce and have 400calories in an ounce.
    Step 1:Frame the system of linear equations based on given conditions
    Let the no. of ounces of candy be x and no. of ounces of cookies be y
    Total cost = cost of x ounces candy + cost of y ounces cookies =
    cross timescost of each ounce candy + y cross times cost of each ounce cookies = $29
    not stretchy rightwards double arrow x space cross times space straight $ 2 plus y space cross times space straight $ 3 equals straight $ 29 not stretchy rightwards double arrow 2 x plus 3 y equals 29— Eq1
    Total no. of calories = calories in x ounces candy + calories in y ounces cookies =
    cross timescalories in each ounce candy + y cross timescalories in each ounce cookies = 4,800
    not stretchy rightwards double arrow x space cross times space 500 plus y space cross times space 400 equals 4 comma 800 not stretchy rightwards double arrow 5 x plus 4 y equals 48 — Eq2
    Step 2:- eliminate x to find the value of y .
    Doing 5(Eq1) -2(Eq2)  to eliminate x .
    5 left parenthesis 2 x plus 3 y right parenthesis minus 2 left parenthesis 5 x plus 4 y right parenthesis equals 5 left parenthesis 29 right parenthesis minus 2 left parenthesis 48 right parenthesis
    10 x minus 10 x plus 15 y minus 8 y equals 145 minus 96
    7 y equals 49
    ∴y = 7
    Step 3:- find x by substituting the value of y in eq1.
    2 x plus 3 y equals 29 not stretchy rightwards double arrow 2 x plus 21 equals 29
    not stretchy rightwards double arrow 2 x equals 29 minus 21 equals 8
    ∴x =4
    ∴ we get the no. of ounces of candy be x = 4 ounces.

    Candies  cost $2 an ounce and have 500 calories in an ounce. Cookies cost $3 an ounce and have 400calories in an ounce. A mixture of candies and cookies costs $29 and has 4,800 calories. Create a system of equations to model this scenario. How many ounces of candies are there in the mixture?

    Maths-General
    Ans :-The no. of ounces of candy is 4 ounces in the mixture .
    Explanation :-
    Given ,candies  cost $2 an ounce and have 500 calories in an ounce. Cookies cost $3 an ounce and have 400calories in an ounce.
    Step 1:Frame the system of linear equations based on given conditions
    Let the no. of ounces of candy be x and no. of ounces of cookies be y
    Total cost = cost of x ounces candy + cost of y ounces cookies =
    cross timescost of each ounce candy + y cross times cost of each ounce cookies = $29
    not stretchy rightwards double arrow x space cross times space straight $ 2 plus y space cross times space straight $ 3 equals straight $ 29 not stretchy rightwards double arrow 2 x plus 3 y equals 29— Eq1
    Total no. of calories = calories in x ounces candy + calories in y ounces cookies =
    cross timescalories in each ounce candy + y cross timescalories in each ounce cookies = 4,800
    not stretchy rightwards double arrow x space cross times space 500 plus y space cross times space 400 equals 4 comma 800 not stretchy rightwards double arrow 5 x plus 4 y equals 48 — Eq2
    Step 2:- eliminate x to find the value of y .
    Doing 5(Eq1) -2(Eq2)  to eliminate x .
    5 left parenthesis 2 x plus 3 y right parenthesis minus 2 left parenthesis 5 x plus 4 y right parenthesis equals 5 left parenthesis 29 right parenthesis minus 2 left parenthesis 48 right parenthesis
    10 x minus 10 x plus 15 y minus 8 y equals 145 minus 96
    7 y equals 49
    ∴y = 7
    Step 3:- find x by substituting the value of y in eq1.
    2 x plus 3 y equals 29 not stretchy rightwards double arrow 2 x plus 21 equals 29
    not stretchy rightwards double arrow 2 x equals 29 minus 21 equals 8
    ∴x =4
    ∴ we get the no. of ounces of candy be x = 4 ounces.
    General
    Maths-

    Chocolate  costs $7 a pound. Biscuits cost $9 a pound. There is a mixture of peanuts and cashews that weighs 10pounds and costs $84. How many pounds of chocolates are there in the mixture?

    Ans :- weight of chocolate is 3 (Pounds) in the mixture of chocolates and biscuits of 10 pounds.
    Explanation :-
    Chocolate cost $7 a pound. Biscuits cost $9 a pound
    Step 1:- frame the system of linear equations with given conditions.
    let the weight of chocolate be x (Pounds) and the weight of biscuits be y(Pounds).
    Given total weight  = weight of chocolates + weight of biscuits  = 10(Pounds)
    x + y = 10 — eq1
    Given total cost of  10 pound mixture  = cost of  x pounds chocolate  + cost of y pounds Biscuits  = $94.
    Cost of x pounds chocolate = x cross times cost of chocolate for 1 pound( = $7 given)
    Cost of x pounds chocolate = x cross times $7 = 7x(in $)
    Cost of y pounds of biscuits = y cross times cost of biscuits for 1 pound  (= $9 given )
    Cost of y pounds of biscuits = y cross times $9 = 9y (in $)
    So total cost of 10 pounds mixture = 7 x plus 9 y equals straight $ 84 not stretchy rightwards double arrow 7 x plus 9 y equals 84 — eq2
    Step 2:- eliminate the x and find y
    Doing eq2 - 3(eq1) to eliminate x .
    7 x plus 9 y minus 7 left parenthesis x plus y right parenthesis equals 84 minus 7 left parenthesis 10 right parenthesis
    not stretchy rightwards double arrow 7 x minus 7 x plus 9 y minus 7 y equals 14
    not stretchy rightwards double arrow 2 y equals 14
    ∴y = 7
    Step 3:- find x by substituting the value of y in eq1.
    x plus y equals 10
    not stretchy rightwards double arrow x plus 7 equals 10
    rightwards double arrow x space equals space 10 space minus space 7
    ∴x = 3
    ∴weight of chocolate is 3 (Pounds) in the mixture of chocolates and biscuits of 10 pounds.

    Chocolate  costs $7 a pound. Biscuits cost $9 a pound. There is a mixture of peanuts and cashews that weighs 10pounds and costs $84. How many pounds of chocolates are there in the mixture?

    Maths-General
    Ans :- weight of chocolate is 3 (Pounds) in the mixture of chocolates and biscuits of 10 pounds.
    Explanation :-
    Chocolate cost $7 a pound. Biscuits cost $9 a pound
    Step 1:- frame the system of linear equations with given conditions.
    let the weight of chocolate be x (Pounds) and the weight of biscuits be y(Pounds).
    Given total weight  = weight of chocolates + weight of biscuits  = 10(Pounds)
    x + y = 10 — eq1
    Given total cost of  10 pound mixture  = cost of  x pounds chocolate  + cost of y pounds Biscuits  = $94.
    Cost of x pounds chocolate = x cross times cost of chocolate for 1 pound( = $7 given)
    Cost of x pounds chocolate = x cross times $7 = 7x(in $)
    Cost of y pounds of biscuits = y cross times cost of biscuits for 1 pound  (= $9 given )
    Cost of y pounds of biscuits = y cross times $9 = 9y (in $)
    So total cost of 10 pounds mixture = 7 x plus 9 y equals straight $ 84 not stretchy rightwards double arrow 7 x plus 9 y equals 84 — eq2
    Step 2:- eliminate the x and find y
    Doing eq2 - 3(eq1) to eliminate x .
    7 x plus 9 y minus 7 left parenthesis x plus y right parenthesis equals 84 minus 7 left parenthesis 10 right parenthesis
    not stretchy rightwards double arrow 7 x minus 7 x plus 9 y minus 7 y equals 14
    not stretchy rightwards double arrow 2 y equals 14
    ∴y = 7
    Step 3:- find x by substituting the value of y in eq1.
    x plus y equals 10
    not stretchy rightwards double arrow x plus 7 equals 10
    rightwards double arrow x space equals space 10 space minus space 7
    ∴x = 3
    ∴weight of chocolate is 3 (Pounds) in the mixture of chocolates and biscuits of 10 pounds.
    General
    Maths-

    A fence is put around a rectangular plot of land. The perimeter of the fence is 28 feet. Two of the opposite sides of the fence cost $10 per foot. The other two sides cost $12 per foot. If the total cost of the fence is $148, what are the dimensions of the fence?

    Ans :- 10 by 4 is the dimensions of the rectangular plot fence
    Explanation :-
    Step 1:- frame the system of linear equations from given conditions
    let the length of plot  be 1 and let the  breadth of plot  be b
    Given total perimeter of the plot   = 2(1 + b)  = 28 not stretchy rightwards double arrow l plus b equals 14  — eq1
    let the breadth b fence cost $12  per foot .
    let the length 1 fence cost $10 per foot
    Given total cost of fence = cost of  fencing length 1 + cost of fencing breadth b
    cost of  fencing length 1 = 1 cross times cost of fencing per foot .
    cost of  fencing length 1   = $10(1)
    cost of  fencing breadth b =cross times b cross timescost of fencing per foot
    cost of  fencing breadth b  = $12(b)
    Total cost of fencing be  = 10(1) + 12b
    10l + 12b = 148 rightwards double arrow5l + 6b = 74 — eq 2
    Step 2 :- eliminate x to find value of y
    Doing  5cross timeseq1 - eq2 to eliminate x
    We get 5l + 5b = (5l + 6b) = 5(14) = 74
    rightwards double arrow5l + 5b - 5l - 6b = 70 - 74
    rightwards double arrow -b = -4
    ∴ b = 4
    Step 3:- substitute the value of y in eq1 to get the value of x
    l plus b equals 14 not stretchy rightwards double arrow l plus 4 equals 14
    not stretchy rightwards double arrow l equals 14 minus 4
    ∴ l = 10.
    ∴ we get the dimensions of rectangular plot = 10 by 4 (in foots)

    A fence is put around a rectangular plot of land. The perimeter of the fence is 28 feet. Two of the opposite sides of the fence cost $10 per foot. The other two sides cost $12 per foot. If the total cost of the fence is $148, what are the dimensions of the fence?

    Maths-General
    Ans :- 10 by 4 is the dimensions of the rectangular plot fence
    Explanation :-
    Step 1:- frame the system of linear equations from given conditions
    let the length of plot  be 1 and let the  breadth of plot  be b
    Given total perimeter of the plot   = 2(1 + b)  = 28 not stretchy rightwards double arrow l plus b equals 14  — eq1
    let the breadth b fence cost $12  per foot .
    let the length 1 fence cost $10 per foot
    Given total cost of fence = cost of  fencing length 1 + cost of fencing breadth b
    cost of  fencing length 1 = 1 cross times cost of fencing per foot .
    cost of  fencing length 1   = $10(1)
    cost of  fencing breadth b =cross times b cross timescost of fencing per foot
    cost of  fencing breadth b  = $12(b)
    Total cost of fencing be  = 10(1) + 12b
    10l + 12b = 148 rightwards double arrow5l + 6b = 74 — eq 2
    Step 2 :- eliminate x to find value of y
    Doing  5cross timeseq1 - eq2 to eliminate x
    We get 5l + 5b = (5l + 6b) = 5(14) = 74
    rightwards double arrow5l + 5b - 5l - 6b = 70 - 74
    rightwards double arrow -b = -4
    ∴ b = 4
    Step 3:- substitute the value of y in eq1 to get the value of x
    l plus b equals 14 not stretchy rightwards double arrow l plus 4 equals 14
    not stretchy rightwards double arrow l equals 14 minus 4
    ∴ l = 10.
    ∴ we get the dimensions of rectangular plot = 10 by 4 (in foots)
    parallel
    General
    Maths-

    A Stone of volume 36 cubic cm is placed in a rectangular tank measuring 8 cm by 6 cm by 5 cm. A student wants to fill the tank with water to a height of 4 cm. What is the volume of the water needed?

    Hint:
    We find the partial volume of the tank with height of the water level given. Then we subtract the volume of stone from the calculated volume of water to get the volume of water needed.
    Explanations:
    Step 1 of 2:
    The length and width of the rectangular tank are given by, 8 cm(= l ) and 6 cm(= b).
    A student wants to fill the tank with water to a height of 4 cm(= h).
    Then, the volume of water needed without the stone is l b h equals 8 cross times 6 cross times 4 equals 192 cm3
    Step 2 of 2:
    The volume of the stone is given by, 36 cm3
    Therefore, the water needed to fill up to desired height is 192 – 36 = 156 cm3
    Final Answer:
    The water needed is 156 cm3.

    A Stone of volume 36 cubic cm is placed in a rectangular tank measuring 8 cm by 6 cm by 5 cm. A student wants to fill the tank with water to a height of 4 cm. What is the volume of the water needed?

    Maths-General
    Hint:
    We find the partial volume of the tank with height of the water level given. Then we subtract the volume of stone from the calculated volume of water to get the volume of water needed.
    Explanations:
    Step 1 of 2:
    The length and width of the rectangular tank are given by, 8 cm(= l ) and 6 cm(= b).
    A student wants to fill the tank with water to a height of 4 cm(= h).
    Then, the volume of water needed without the stone is l b h equals 8 cross times 6 cross times 4 equals 192 cm3
    Step 2 of 2:
    The volume of the stone is given by, 36 cm3
    Therefore, the water needed to fill up to desired height is 192 – 36 = 156 cm3
    Final Answer:
    The water needed is 156 cm3.
    General
    Maths-

    The area of a trapezium side field is 480sq. m , the height is 15m and one of the parallel side is 20m .Find the other side ?

    Ans :- 44 m.
    Explanation :-
    Given, area of trapezium = 480 sq.m ,the length of  parallel side is 20 m
    Let the length of the other parallel side be x m .
    Altitude or height = 15 m
    text  Area of trapezium  end text equals 1 half left parenthesis text  height  end text right parenthesis left parenthesis text  sum of lengths of parallel sides  end text right parenthesis
    not stretchy rightwards double arrow 480 equals 1 half left parenthesis 15 right parenthesis left parenthesis 20 plus x right parenthesis not stretchy rightwards double arrow 480 cross times 2 over 15 equals 20 plus x not stretchy rightwards double arrow 64 equals 20 plus x
    not stretchy rightwards double arrow x equals 64 minus 20 not stretchy rightwards double arrow x equals 44
    Therefore, the length of the unknown parallel side is 44 m.

    The area of a trapezium side field is 480sq. m , the height is 15m and one of the parallel side is 20m .Find the other side ?

    Maths-General
    Ans :- 44 m.
    Explanation :-
    Given, area of trapezium = 480 sq.m ,the length of  parallel side is 20 m
    Let the length of the other parallel side be x m .
    Altitude or height = 15 m
    text  Area of trapezium  end text equals 1 half left parenthesis text  height  end text right parenthesis left parenthesis text  sum of lengths of parallel sides  end text right parenthesis
    not stretchy rightwards double arrow 480 equals 1 half left parenthesis 15 right parenthesis left parenthesis 20 plus x right parenthesis not stretchy rightwards double arrow 480 cross times 2 over 15 equals 20 plus x not stretchy rightwards double arrow 64 equals 20 plus x
    not stretchy rightwards double arrow x equals 64 minus 20 not stretchy rightwards double arrow x equals 44
    Therefore, the length of the unknown parallel side is 44 m.
    General
    Maths-

    The velocity v , in meters per second, of a falling object on Earth after t seconds, ignoring the effect of air resistance, is modeled by the equation v = 9.8t . There is a different linear relationship between time and velocity on Mars, as shown in the table below.

    If an object dropped toward the surface of Earth has a velocity of  58.8 meters per second after t  seconds, what would be the velocity of the same object dropped toward the surface of Mars after t  seconds, ignoring the effect of air resistance?

    First we find the equation of velocity of a falling object on Mars (ignoring the effect of air resistance).
    It is given that the relationship is linear, so the ratio of velocity and time is constant.
    From the table, we have
    v over t equals fraction numerator 14.8 over denominator 4 end fraction equals fraction numerator 29.6 over denominator 8 end fraction equals 3.7
    Thus, the equation of velocity with respect to time of a falling object on Mars is  v = 3.7t
    Given, equation of velocity of a falling object on Earth with respect to time is v = 9.8t
    According to the question, an object dropped toward the surface of Earth has a velocity of 58.8 meters per second after t seconds
    By the equation v = 9.8t , the time after which the velocity is 58.8 metres per second is given by
    t equals fraction numerator 58.8 over denominator 9.8 end fraction equals 6 text  second  end text s
    We need to find the velocity of the same object dropped towards the surface of mass after t=6 seconds
    By the relation v = 3.7t , we get the velocity to be
    v equals 3.7 cross times 6 equals 22.2 text  metres per second  end text
    Thus, the correct option is B)

    The velocity v , in meters per second, of a falling object on Earth after t seconds, ignoring the effect of air resistance, is modeled by the equation v = 9.8t . There is a different linear relationship between time and velocity on Mars, as shown in the table below.

    If an object dropped toward the surface of Earth has a velocity of  58.8 meters per second after t  seconds, what would be the velocity of the same object dropped toward the surface of Mars after t  seconds, ignoring the effect of air resistance?

    Maths-General
    First we find the equation of velocity of a falling object on Mars (ignoring the effect of air resistance).
    It is given that the relationship is linear, so the ratio of velocity and time is constant.
    From the table, we have
    v over t equals fraction numerator 14.8 over denominator 4 end fraction equals fraction numerator 29.6 over denominator 8 end fraction equals 3.7
    Thus, the equation of velocity with respect to time of a falling object on Mars is  v = 3.7t
    Given, equation of velocity of a falling object on Earth with respect to time is v = 9.8t
    According to the question, an object dropped toward the surface of Earth has a velocity of 58.8 meters per second after t seconds
    By the equation v = 9.8t , the time after which the velocity is 58.8 metres per second is given by
    t equals fraction numerator 58.8 over denominator 9.8 end fraction equals 6 text  second  end text s
    We need to find the velocity of the same object dropped towards the surface of mass after t=6 seconds
    By the relation v = 3.7t , we get the velocity to be
    v equals 3.7 cross times 6 equals 22.2 text  metres per second  end text
    Thus, the correct option is B)
    parallel
    General
    Maths-

    A pet store has 30 animals. Some are cats and the rest are dogs. The cats cost $50 each. The dogs cost $100each. If the total cost for all 30 animals is $1900, how many cats are there?

    Ans :-The animal shop has 22 cats .
    Explanation :-
    Step 1:- construct the system of linear equations
    let the no. of cats be x and let the no. of dogs  be y
    Given total no. of animals   = no. of cats   + no. of dogs  = 30
    So, x + y = 30 — eq1
    Given total cost of all animals  = cost of x cats  + cost of y dogs = $1900.
    cost of x cats = x cross times cost of each cat ( The cats cost $50 each)
    cost of x cats  = 50x (in $ )
    cost of y dogs = y cross timescost of each dog ( The dogs cost $100each)
    cost of y dogs  = 100y (in $)
    So total cost of all animals = 50x + 100y = 1.900 — eq2
    Step 2 :- eliminate x to find value of y
    Doing  eq2 - 150 cross timeseq1 to eliminate x
    We get 50x + 100y - 50(x + y) = 1,900 - 50(30)
    not stretchy rightwards double arrow 100 y minus 50 y equals 1 comma 900 minus 1 comma 500
    not stretchy rightwards double arrow 50 y equals 400 not stretchy rightwards double arrow y equals 8
    ∴ y = 8
    Step 3:- substitute the value of y in eq1 to get the value of x
    x plus y equals 30 not stretchy rightwards double arrow x plus 8 equals 30
    not stretchy rightwards double arrow x equals 30 minus 8
    ∴ x = 22
    ∴ The animals shop has 22 cats and 8 dogs in total of 30 animals.

    A pet store has 30 animals. Some are cats and the rest are dogs. The cats cost $50 each. The dogs cost $100each. If the total cost for all 30 animals is $1900, how many cats are there?

    Maths-General
    Ans :-The animal shop has 22 cats .
    Explanation :-
    Step 1:- construct the system of linear equations
    let the no. of cats be x and let the no. of dogs  be y
    Given total no. of animals   = no. of cats   + no. of dogs  = 30
    So, x + y = 30 — eq1
    Given total cost of all animals  = cost of x cats  + cost of y dogs = $1900.
    cost of x cats = x cross times cost of each cat ( The cats cost $50 each)
    cost of x cats  = 50x (in $ )
    cost of y dogs = y cross timescost of each dog ( The dogs cost $100each)
    cost of y dogs  = 100y (in $)
    So total cost of all animals = 50x + 100y = 1.900 — eq2
    Step 2 :- eliminate x to find value of y
    Doing  eq2 - 150 cross timeseq1 to eliminate x
    We get 50x + 100y - 50(x + y) = 1,900 - 50(30)
    not stretchy rightwards double arrow 100 y minus 50 y equals 1 comma 900 minus 1 comma 500
    not stretchy rightwards double arrow 50 y equals 400 not stretchy rightwards double arrow y equals 8
    ∴ y = 8
    Step 3:- substitute the value of y in eq1 to get the value of x
    x plus y equals 30 not stretchy rightwards double arrow x plus 8 equals 30
    not stretchy rightwards double arrow x equals 30 minus 8
    ∴ x = 22
    ∴ The animals shop has 22 cats and 8 dogs in total of 30 animals.
    General
    Maths-

    The parallel sides of a trapezium are 30 cm and 10 cm and the other two sides are 13 cm and 21 cm. Find its area?

    Ans :- 252 cm2.
    Explanation :-
    Step 1:- Given lengths of parallel sides is 10 and 30 cm .
    Length of non parallel sides is 21 and 13 cm
    We get a triangle and parallelogram ABED
    we get BE = 21 cm ; DE = 10 cm (opposites sides of parallelogram )
    CE = CD-DE = 30 - 10 = 20 cm .


    Step 2:- Equate the areas and find value of height h
    10 h equals 126 not stretchy rightwards double arrow h equals 126 over 10 equals 12.6 cm
    Step 3:-Find the area of trapezium
    text  Area of trapezium  end text equals 1 half cross times left parenthesis text  height  end text right parenthesis cross times left parenthesis text  sum of lengths of parallelsides)  end text
    text  Area of trapezium  end text equals 1 half open parentheses bold 126 over bold 10 close parentheses left parenthesis 10 plus 30 right parenthesis equals 126 cross times 2 equals 252 cm squared
    Therefore, Area of trapezium ABCD  = 252 cm2.

    The parallel sides of a trapezium are 30 cm and 10 cm and the other two sides are 13 cm and 21 cm. Find its area?

    Maths-General
    Ans :- 252 cm2.
    Explanation :-
    Step 1:- Given lengths of parallel sides is 10 and 30 cm .
    Length of non parallel sides is 21 and 13 cm
    We get a triangle and parallelogram ABED
    we get BE = 21 cm ; DE = 10 cm (opposites sides of parallelogram )
    CE = CD-DE = 30 - 10 = 20 cm .


    Step 2:- Equate the areas and find value of height h
    10 h equals 126 not stretchy rightwards double arrow h equals 126 over 10 equals 12.6 cm
    Step 3:-Find the area of trapezium
    text  Area of trapezium  end text equals 1 half cross times left parenthesis text  height  end text right parenthesis cross times left parenthesis text  sum of lengths of parallelsides)  end text
    text  Area of trapezium  end text equals 1 half open parentheses bold 126 over bold 10 close parentheses left parenthesis 10 plus 30 right parenthesis equals 126 cross times 2 equals 252 cm squared
    Therefore, Area of trapezium ABCD  = 252 cm2.

    General
    Maths-

    A Hollow cylindrical pipe is 21 cm long. If its outer and inner diameters are 10 cm and 6 cm respectively, then find the volume of the metal used in making the pipe?

    Hint:
    We find the inner and outer radius of the pipe. Then calculate the volume of the pipe using both the radius. At last, we subtract the volume with inner radius from the volume with outer radius to find the volume of metal.
    Explanations:
    Step 1 of 3:
    The outer and inner diameters of the hollow cylindrical pipe are given by, 10 cm and 6 cm.
    Then, the outer and inner radius are 5 cm(= R) and 3 cm(= r).
    The pipe is 21 cm(= h) long.
    Step 2 of 3:
    The volume with outer radius is pi R squared h equals 22 over 7 cross times 5 squared cross times 21 equals 1650 cm3
    The volume with inner radius is pi r squared h equals 22 over 7 cross times 3 squared cross times 21 equals 594 cm3
    Step 3 of 3:
    Volume of metal used to make the pipe is pi R squared h minus pi r squared h equals 1650 minus 594 equals 1056 cm3
    Final Answer:
    The volume of metal used in making the pipe is 1056 cm3.

    A Hollow cylindrical pipe is 21 cm long. If its outer and inner diameters are 10 cm and 6 cm respectively, then find the volume of the metal used in making the pipe?

    Maths-General
    Hint:
    We find the inner and outer radius of the pipe. Then calculate the volume of the pipe using both the radius. At last, we subtract the volume with inner radius from the volume with outer radius to find the volume of metal.
    Explanations:
    Step 1 of 3:
    The outer and inner diameters of the hollow cylindrical pipe are given by, 10 cm and 6 cm.
    Then, the outer and inner radius are 5 cm(= R) and 3 cm(= r).
    The pipe is 21 cm(= h) long.
    Step 2 of 3:
    The volume with outer radius is pi R squared h equals 22 over 7 cross times 5 squared cross times 21 equals 1650 cm3
    The volume with inner radius is pi r squared h equals 22 over 7 cross times 3 squared cross times 21 equals 594 cm3
    Step 3 of 3:
    Volume of metal used to make the pipe is pi R squared h minus pi r squared h equals 1650 minus 594 equals 1056 cm3
    Final Answer:
    The volume of metal used in making the pipe is 1056 cm3.
    parallel
    General
    Maths-

    There are 8 people in the lift. Few are ladies and the rest are children. Each lady weighs 150 pounds. Each child weighs 50 pounds. The total weight of the 8 people is 800 pounds. Construct the system of linear equations and use elimination method to solve.

    Ans :- The no. of ladies in the lift are 4 and no. of children in the lift are 4 .
    Explanation :-
    Step 1:- construct the system of linear equations
    let the no. of ladies in the lift be x and let the number of children  be y
    Given total no. of people  = no. of ladies  + no. of children  = 8
    So, x + y = 8 — eq1
    Given total weight  = weight of x ladies + weight of y children = 800.
    weight of x ladies = x cross times weight of each lady ( Each lady weighs 150 pounds)
    weight of x ladies = 150x (pounds)
    weight of y children = y cross times weight of each child ( Each child weighs 50 pounds)
    weight of y children  = 50y (pounds)
    So total weight = 150x + 50y = 800 — eq2
    Step 2 :- eliminate x to find value of y
    Doing  eq2 - 150 cross timeseq1 to eliminate x
    We get 150x + 50y - 150 (x + y) = 800 - 150(8)
    not stretchy rightwards double arrow 50 y minus 150 y equals 800 minus 1200
    not stretchy rightwards double arrow negative 100 y equals negative 400 not stretchy rightwards double arrow y equals 4
    ∴ y = 4
    Step 3:- substitute the value of y in eq1 to get the value of x
    x plus y equals 8 not stretchy rightwards double arrow x plus 4 equals 8
    not stretchy rightwards double arrow x equals 8 minus 4
    ∴ x = 4
    ∴The no. of ladies in the lift are 4 and no. of children in the lift are 4 .

    There are 8 people in the lift. Few are ladies and the rest are children. Each lady weighs 150 pounds. Each child weighs 50 pounds. The total weight of the 8 people is 800 pounds. Construct the system of linear equations and use elimination method to solve.

    Maths-General
    Ans :- The no. of ladies in the lift are 4 and no. of children in the lift are 4 .
    Explanation :-
    Step 1:- construct the system of linear equations
    let the no. of ladies in the lift be x and let the number of children  be y
    Given total no. of people  = no. of ladies  + no. of children  = 8
    So, x + y = 8 — eq1
    Given total weight  = weight of x ladies + weight of y children = 800.
    weight of x ladies = x cross times weight of each lady ( Each lady weighs 150 pounds)
    weight of x ladies = 150x (pounds)
    weight of y children = y cross times weight of each child ( Each child weighs 50 pounds)
    weight of y children  = 50y (pounds)
    So total weight = 150x + 50y = 800 — eq2
    Step 2 :- eliminate x to find value of y
    Doing  eq2 - 150 cross timeseq1 to eliminate x
    We get 150x + 50y - 150 (x + y) = 800 - 150(8)
    not stretchy rightwards double arrow 50 y minus 150 y equals 800 minus 1200
    not stretchy rightwards double arrow negative 100 y equals negative 400 not stretchy rightwards double arrow y equals 4
    ∴ y = 4
    Step 3:- substitute the value of y in eq1 to get the value of x
    x plus y equals 8 not stretchy rightwards double arrow x plus 4 equals 8
    not stretchy rightwards double arrow x equals 8 minus 4
    ∴ x = 4
    ∴The no. of ladies in the lift are 4 and no. of children in the lift are 4 .
    General
    Maths-

    If the radius and height of a cylinder are in ratio 5 :7 and its volume is 550 cubic.cm, then find the radius?

    Hint:
    We use the common ratio to find the radius. Moreover, we would only use the positive of the cube root since radius is a distance and distance cannot be negative.
    Explanations:
    Step 1 of 3:
    The radius and height of a cylinder are in ratio 5:7.
    Let the common ratio be x.
    Then, the radius and height of the cylinder is given by, 5x(= r) and 7x(= h) unit, respectively.
    Step 2 of 3:
    text  Given,  end text pi r squared h equals 550
    not stretchy rightwards double arrow 22 over 7 cross times left parenthesis 5 x right parenthesis squared cross times 7 x equals 550
    not stretchy rightwards double arrow 22 over 7 cross times 25 x squared cross times 7 x equals 550

    not stretchy rightwards double arrow 22 over 7 cross times 175 x cubed equals 550
    not stretchy rightwards double arrow x cubed equals 550 cross times 7 over 22 cross times 1 over 175
    not stretchy rightwards double arrow x cubed equals 1
    not stretchy rightwards double arrow x equals 1
    Step 3 of 3:
    The common ratio is x = 1.
    Therefore, the radius of the cylinder is 5 x equals 5 cross times 1 equals 5 cm
    Final Answer:
    The radius is 5 cm.

    If the radius and height of a cylinder are in ratio 5 :7 and its volume is 550 cubic.cm, then find the radius?

    Maths-General
    Hint:
    We use the common ratio to find the radius. Moreover, we would only use the positive of the cube root since radius is a distance and distance cannot be negative.
    Explanations:
    Step 1 of 3:
    The radius and height of a cylinder are in ratio 5:7.
    Let the common ratio be x.
    Then, the radius and height of the cylinder is given by, 5x(= r) and 7x(= h) unit, respectively.
    Step 2 of 3:
    text  Given,  end text pi r squared h equals 550
    not stretchy rightwards double arrow 22 over 7 cross times left parenthesis 5 x right parenthesis squared cross times 7 x equals 550
    not stretchy rightwards double arrow 22 over 7 cross times 25 x squared cross times 7 x equals 550

    not stretchy rightwards double arrow 22 over 7 cross times 175 x cubed equals 550
    not stretchy rightwards double arrow x cubed equals 550 cross times 7 over 22 cross times 1 over 175
    not stretchy rightwards double arrow x cubed equals 1
    not stretchy rightwards double arrow x equals 1
    Step 3 of 3:
    The common ratio is x = 1.
    Therefore, the radius of the cylinder is 5 x equals 5 cross times 1 equals 5 cm
    Final Answer:
    The radius is 5 cm.
    General
    Maths-


    Which of the following is true about the standard deviations of the two data sets in the table above?

    The mean of data set A is given by:
    mu equals left parenthesis 25550 plus 40430 plus 49150 plus 62590 plus 73670 plus 118780 plus 126040 right parenthesis divided by 7
    equals 496210 divided by 7
    equals 70887.14
    The mean of data set B is given by:
    mu equals left parenthesis 22860 plus 55020 plus 173730 plus 300580 plus 358920 plus 456170 plus 603300 right parenthesis divided by 7
    equals 281511.42
    We can clearly observe that the values in Data set B are dispersed from the mean, whereas the values in Data set A are not so dispersed.
    Hence, the standard deviation of data set B is larger than the standard deviation of data set A.
    Thus, the correct option is A


    Which of the following is true about the standard deviations of the two data sets in the table above?

    Maths-General
    The mean of data set A is given by:
    mu equals left parenthesis 25550 plus 40430 plus 49150 plus 62590 plus 73670 plus 118780 plus 126040 right parenthesis divided by 7
    equals 496210 divided by 7
    equals 70887.14
    The mean of data set B is given by:
    mu equals left parenthesis 22860 plus 55020 plus 173730 plus 300580 plus 358920 plus 456170 plus 603300 right parenthesis divided by 7
    equals 281511.42
    We can clearly observe that the values in Data set B are dispersed from the mean, whereas the values in Data set A are not so dispersed.
    Hence, the standard deviation of data set B is larger than the standard deviation of data set A.
    Thus, the correct option is A
    parallel
    General
    Maths-

    50 animals are there in a shed. Some of the animals have 2 legs and the rest of them have 4 legs. In total there are 172 legs. Frame the system of linear equations and use elimination method to solve .

    Ans :- The no. of 2 leg animals are 26 and no. of 4 leg animals are 24.
    Explanation :-
    Step 1:- Frame the equations
    let the no. of 2 leg animals be x and let the number of 4 leg animals be y
    Given total no. of animals = no. of 2 leg animals + no. of 4 leg animals = 50
    i. e x + y = 50 — eq1
    Given no. of legs = 2cross timesno. of 2 leg animals  + 4cross timesno. of 4 leg animals = 172.
    i. e 2x+4y = 172 — eq2
    So the equations are x + y = 50 and 2x + 4y = 172
    Step 2 :- eliminate x to find value of y
    Doing  eq2 - 2cross timeseq1 to eliminate x
    We get 2 x plus 4 v minus 2 left parenthesis x plus y right parenthesis equals 172 minus 2 left parenthesis 50 right parenthesis
    not stretchy rightwards double arrow 4 y minus y equals 172 minus 100
    not stretchy rightwards double arrow 3 y equals 72
    ∴ y = 24
    Step 3:- substitute the value of y in eq1 to get the value of x
    x plus y equals 50 not stretchy rightwards double arrow x plus 24 equals 50
    not stretchy rightwards double arrow x equals 50 minus 24
    ∴ x = 26
    ∴The no. of 2 leg animals are 26 and no. of 4 leg animals are 24.

    50 animals are there in a shed. Some of the animals have 2 legs and the rest of them have 4 legs. In total there are 172 legs. Frame the system of linear equations and use elimination method to solve .

    Maths-General
    Ans :- The no. of 2 leg animals are 26 and no. of 4 leg animals are 24.
    Explanation :-
    Step 1:- Frame the equations
    let the no. of 2 leg animals be x and let the number of 4 leg animals be y
    Given total no. of animals = no. of 2 leg animals + no. of 4 leg animals = 50
    i. e x + y = 50 — eq1
    Given no. of legs = 2cross timesno. of 2 leg animals  + 4cross timesno. of 4 leg animals = 172.
    i. e 2x+4y = 172 — eq2
    So the equations are x + y = 50 and 2x + 4y = 172
    Step 2 :- eliminate x to find value of y
    Doing  eq2 - 2cross timeseq1 to eliminate x
    We get 2 x plus 4 v minus 2 left parenthesis x plus y right parenthesis equals 172 minus 2 left parenthesis 50 right parenthesis
    not stretchy rightwards double arrow 4 y minus y equals 172 minus 100
    not stretchy rightwards double arrow 3 y equals 72
    ∴ y = 24
    Step 3:- substitute the value of y in eq1 to get the value of x
    x plus y equals 50 not stretchy rightwards double arrow x plus 24 equals 50
    not stretchy rightwards double arrow x equals 50 minus 24
    ∴ x = 26
    ∴The no. of 2 leg animals are 26 and no. of 4 leg animals are 24.
    General
    Maths-

    The area of trapezium is 384 sq.cm. If its parallel sides are in the ratio 3:5 and the perpendicular distance between them 12cm. The smaller of the parallel side is

    Ans :- 24 cm
    Explanation :-
    Step 1:-  Find the equation between x and y  and find the sum of lengths of parallel sides in terms of y.
    let the length of parallel sides be x and y (x<y).
    Given ratio of x:y = 3:5 not stretchy rightwards double arrow x over y equals 3 over 5 not stretchy rightwards double arrow y equals 5 x divided by 3
    Sum of lengths of parallel sides = x plus y equals x plus 5 x divided by 3 equals 8 x divided by 3
    Step 2:-substitute the values in the area formula find x
    Given area of trapezium = 384 sq.cm
    Altitude (or) height = 12 cm
    area of trapezium = ½ × (height)(sum of lengths of parallel side)
    not stretchy rightwards double arrow 384 equals 1 half cross times left parenthesis 12 right parenthesis cross times open parentheses fraction numerator 8 x over denominator 3 end fraction close parentheses not stretchy rightwards double arrow 384 equals 2 cross times 8 x not stretchy rightwards double arrow x equals 384 over 16
    not stretchy rightwards double arrow x equals 24 cm
    We  get x = 24 cm as x is the smaller side of parallel sides(x<y)
    Therefore, the length of smaller parallel sides is 24 cm .

    The area of trapezium is 384 sq.cm. If its parallel sides are in the ratio 3:5 and the perpendicular distance between them 12cm. The smaller of the parallel side is

    Maths-General
    Ans :- 24 cm
    Explanation :-
    Step 1:-  Find the equation between x and y  and find the sum of lengths of parallel sides in terms of y.
    let the length of parallel sides be x and y (x<y).
    Given ratio of x:y = 3:5 not stretchy rightwards double arrow x over y equals 3 over 5 not stretchy rightwards double arrow y equals 5 x divided by 3
    Sum of lengths of parallel sides = x plus y equals x plus 5 x divided by 3 equals 8 x divided by 3
    Step 2:-substitute the values in the area formula find x
    Given area of trapezium = 384 sq.cm
    Altitude (or) height = 12 cm
    area of trapezium = ½ × (height)(sum of lengths of parallel side)
    not stretchy rightwards double arrow 384 equals 1 half cross times left parenthesis 12 right parenthesis cross times open parentheses fraction numerator 8 x over denominator 3 end fraction close parentheses not stretchy rightwards double arrow 384 equals 2 cross times 8 x not stretchy rightwards double arrow x equals 384 over 16
    not stretchy rightwards double arrow x equals 24 cm
    We  get x = 24 cm as x is the smaller side of parallel sides(x<y)
    Therefore, the length of smaller parallel sides is 24 cm .

    General
    Maths-

    A Rectangular sheet of paper 44 cm x 18 cm is rolled along its length and a cylinder is formed. What is the volume of the cylinder so formed?

    Hint:
    If a rectangular sheet of paper is rolled along its length to form a cylinder, then its length becomes the circumference of the cylinder base and its width becomes the height of the cylinder.
    Explanations:
    Step 1 of 3:
    The length and width of a rectangular sheet of paper are given by, 44 cm and 18 cm, respectively.
    As per the hint, if the paper is rolled along its length to form a cylinder, then the height of the cylinder would be 18 cm(= h), and the circumference of the cylinder base would be 44 cm.
    Step 2 of 3:
    Therefore, 2 pi r equals 44 not stretchy rightwards double arrow r equals 44 cross times 7 over 22 cross times 1 half equals 7 cm
    The radius of the cylinder formed is 7 cm(= r).
    Step 3 of 3:
    The volume of the formed cylinder is pi r squared h equals 22 over 7 cross times 7 squared cross times 18 equals 2772 cm3
    Final Answer:
    The volume of the cylinder so formed is 2772 cm3.

    A Rectangular sheet of paper 44 cm x 18 cm is rolled along its length and a cylinder is formed. What is the volume of the cylinder so formed?

    Maths-General
    Hint:
    If a rectangular sheet of paper is rolled along its length to form a cylinder, then its length becomes the circumference of the cylinder base and its width becomes the height of the cylinder.
    Explanations:
    Step 1 of 3:
    The length and width of a rectangular sheet of paper are given by, 44 cm and 18 cm, respectively.
    As per the hint, if the paper is rolled along its length to form a cylinder, then the height of the cylinder would be 18 cm(= h), and the circumference of the cylinder base would be 44 cm.
    Step 2 of 3:
    Therefore, 2 pi r equals 44 not stretchy rightwards double arrow r equals 44 cross times 7 over 22 cross times 1 half equals 7 cm
    The radius of the cylinder formed is 7 cm(= r).
    Step 3 of 3:
    The volume of the formed cylinder is pi r squared h equals 22 over 7 cross times 7 squared cross times 18 equals 2772 cm3
    Final Answer:
    The volume of the cylinder so formed is 2772 cm3.
    parallel

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